An integer b is divisible by an integer a, not zero, if there exists an integer x such that b = ax and we write a|b{\displaystyle a|b}. In case b is not divisible by a, we write a∤b{\displaystyle a\nmid b}. For example 3|6{\displaystyle 3|6} and 4∤6{\displaystyle 4\nmid 6}.

If a|b{\displaystyle a|b} and 0 < a < b then a is called a proper divisor of b. The notation ak||b{\displaystyle a^{k}||b} is used to indicate that ak|b{\displaystyle a^{k}|b} but ak+1∤b{\displaystyle a^{k+1}\nmid b}. For example 3||6{\displaystyle 3||6}.

a|b{\displaystyle a|b} and a|c{\displaystyle a|c} imply existence of m,n such that b = am and c = an so that bx + cy = amx + any = az and so a|bx+cy{\displaystyle a|bx+cy}.

a|b{\displaystyle a|b} and b|a{\displaystyle b|a} imply that m,n exist such that b = am and a = bn. Then a = bn = amn and so mn = 1. The only choice for m and n is to be simultaneously 1 or -1. In either case we have the result.

b = ax for some x > 0 as otherwise a and b would have opposite signs. So b = (a + a + ...) x times ≥{\displaystyle \geq } a.

Proof: The proof consists of two parts — first, the proof of the existence of q and r, and secondly, the proof of the uniqueness of q and r.

Let us first consider existence.

Consider the set

S={a−nd:n∈Z}{\displaystyle S=\left\{a-nd:n\in \mathbb {Z} \right\}}

We claim that S contains at least one nonnegative integer. There are two cases to consider.

If d < 0, then −d > 0, and by the Archimedean property, there is a nonnegative integer n such that (−d)n ≥ −a, i.e. a − dn ≥ 0.

If d > 0, then again by the Archimedean property, there is a nonnegative integer n such that dn ≥ −a, i.e. a − d(−n) = a + dn ≥ 0.

In either case, we have shown that S contains a nonnegative integer. This means we can apply the well-ordering principle, and deduce that S contains a least nonnegative integer r. If we now let q = (a − r)/d, then q and r are integers and a = qd + r.

It only remains to show that 0 ≤ r < |d|. The first inequality holds because of the choice of r as a nonnegative integer. To show the last (strict) inequality, suppose that r ≥ |d|. Since d ≠ 0, r > 0, and again d > 0 or d < 0.

If d > 0, then r ≥ d implies a-qd ≥ d. This implies that a-qd-d ≥0, further implying that a-(q+1)d ≥ 0. Therefore, a-(q+1)d is in S and, since a-(q+1)d=r-d with d>0 we know a-(q+1)d<r, contradicting the assumption that r was the least nonnegative element of S.

If d<0 then r ≥ -d implying that a-qd ≥ -d. This implies that a-qd+d ≥0, further implying that a-(q-1)d ≥ 0. Therefore, a-(q-1)d is in S and, since a-(q-1)d=r+d with d<0 we know a-(q-1)d<r, contradicting the assumption that r was the least nonnegative element of S.

In either case, we have shown that r > 0 was not really the least nonnegative integer in S, after all. This is a contradiction, and so we must have r < |d|. This completes the proof of the existence of q and r.

The original equation implies that |d| divides |r- r' |; therefore either |d| ≤ |r- 'r' | (if |r- r' | > 0 so that |d| is also > 0 and property 5 of Basic Properties above holds), or |r- r' |=0. Because we just established that |r-r' | < |d|, we may conclude that the first possibility cannot hold. Thus, r=r' .

Substituting this into the original two equations quickly yields dq = dq' and, since we assumed d is not 0, it must be the case that q = q' proving uniqueness.

Remarks:

The name division algorithm is something of a misnomer, as it is a theorem, not an algorithm, i.e. a well-defined procedure for achieving a specific task.

There is nothing particularly special about the set of remainders {0, 1, ..., |d| − 1}. We could use any set of |d| integers, such that every integer is congruent to one of the integers in the set. This particular set of remainders is very convenient, but it is not the only choice.